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a(1)=1, a(n) = 6*a(n-1) + n.
6

%I #33 Feb 12 2023 10:31:01

%S 1,8,51,310,1865,11196,67183,403106,2418645,14511880,87071291,

%T 522427758,3134566561,18807399380,112844396295,677066377786,

%U 4062398266733,24374389600416,146246337602515,877478025615110

%N a(1)=1, a(n) = 6*a(n-1) + n.

%H Vincenzo Librandi, <a href="/A014829/b014829.txt">Table of n, a(n) for n = 1..1000</a>

%H László Tóth, <a href="https://arxiv.org/abs/2002.06584">On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers</a>, arXiv:2002.06584 [math.NT], 2020. See also <a href="https://doi.org/10.1090/proc/14863">Proc. Amer. Math. Soc.</a> 148 (2020), 461-469.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (8,-13,6).

%F a(n) = (6^(n+1) - 5*n - 6)/25. - _Rolf Pleisch_, Oct 25 2010

%F Binomial transform of x*(1+x)/(1-5*x), or A003948 with a leading 0. a(n) = Sum_{k=0..n} (n-k)*6^k = Sum_{k=0..n} k*6^(n-k); a(n) = Sum_{k=0..n} binomial(n+2, k+2)*5^k [Offset 0]. - _Paul Barry_, Jul 30 2004

%F From _Colin Barker_, Jun 03 2020: (Start)

%F G.f.: x / ((1 - x)^2*(1 - 6*x)).

%F a(n) = 8*a(n-1) - 13*a(n-2) + 6*a(n-3) for n>3.

%F (End)

%p a:=n->1/5*sum(6^j-1,j=1..n): seq(a(n),n=1..20); # _Zerinvary Lajos_, Jun 27 2007

%t Join[{a=1,b=8},Table[c=7*b-6*a+1;a=b;b=c,{n,60}]] (* _Vladimir Joseph Stephan Orlovsky_, Feb 06 2011 *)

%t nxt[{n_,a_}]:={n+1,6a+n+1}; NestList[nxt,{1,1},30][[All,2]] (* _Harvey P. Dale_, Feb 12 2023 *)

%o (Magma) [(6^(n+1)-5*n-6)/25: n in [1..30]]; // _Vincenzo Librandi_, Aug 23 2011

%o (PARI) Vec(x / ((1 - x)^2*(1 - 6*x)) + O(x^25)) \\ _Colin Barker_, Jun 03 2020

%K nonn,easy

%O 1,2

%A _N. J. A. Sloane_